CMPSCI 190DM: A Mathematical Foundation for Informatics

Solutions to Second Midterm Exam

David Mix Barrington

23 October 2015

Directions:

Answer the problems on the exam pages.
There are four problems for 100 total points. Actual scale was A = 90, C = 60.
If you need extra space use the back of a page.
No books, notes, calculators, or collaboration.

  Q1: 20 points
  Q2: 15 points
  Q3: 20 points
  Q4: 20 points
  Q5: 25 points
Total: 100 points

Question text is in black, solutions in blue.

Question 1 (20): Briefly identify the following terms or concepts (4 points each):

(a) the Pigeonhole Principle
The Pigeonhole Principle says that if a set of mn + 1 or more items is divided into n groups, at least one group has at least m + 1 items.
(b) an odd number
An odd number is an integer that can be written as 2k + 1, where k is another integer.
(c) the Division Theorem
The Division Theorem says that if n is an integer and m a positive integer, there exist integers q and r such that n = qm + r and 0 ≤ r < m.
(d) the mod operation
The mod operation, written "n mod m" in the textbook but (mostly) written "n % m" in Java, returns the number r from the Division Theorem above. It returns the remainder when n is divided by m. (When n is negative, the textbook's mod operation returns r such that 0 ≤ r < m, while Java's % operation returns a negative number.)
(e) a perfect square
A perfect square is an integer that can be written as k² for some integer k.

Question 2 (15): Identify each of the following statements as true or false. No explanation is needed or wanted, and there is no penalty for guessing. (3 points each)

(a) The binary representation of the decimal number 36 is 100010.
FALSE. The correct binary for 36 is 100100.
(b) If S is a set of eleven different positive integers, there exist two elements x and y of S, with x ≠ y, such that x and y have the same last digit in decimal notation.
TRUE. If we divide the 11 numbers into 10 groups based on their last digit, the Pigeonhole Principle says that at least one group must have at least two elements.
(c) The sum of any two rational numbers is rational.
TRUE. If x and y are arbitrary rational numbers, then they can be written as x = a/b and y = c/d, where a and c are integers and b and d are nonzero integers. Then x + y = (ad + bc)/bd, which is an integer divided by a nonzero integer and thus a rational number.
(d) Every number that is divisible by 6 is also divisible by 12.
FALSE. The number 6 is divisible by 6 but not by 12. The converse of this statement is true.
(e) If n is an odd number, then (n-3)³ + 6 must be an even number.
TRUE. If n is odd, then n - 3 is even, its square is also even, and we get another even number when we add the even number 6 to it.

Question 3 (20): Prove that if an integer n is not divisible by 4, then n³ is also not divisible by 4. (Hint: Use Proof by Cases and the Division Theorem on n, with m = 4.)

Following the hint, we write n as 4q + r and consider the four cases for the four possible values of r.

If n = 4q, it is divisible by 4 and we have nothing to prove.

If n = 4q + 1, n³ + 2 = 64q³ + 48q² + 12q + 3, which is not divisible by 4.

If n = 4q + 2, n³ + 2 = 64q³ + 96q² + 48q + 10, which is not divisible by 4.

If n = 4q + 3, n³ + 2 = 64q³ + 144q² + 108q + 29, which is not divisible by 4.

Question 4 (20): Define a number sequence so that a₁ = 3 and, for any n with n > 1, a_n = a_n-1 + 3n² - 3n + 1. Prove by induction that for all positive integers n, a_n = n³ + 2.

Our P(n) is the statement "a_n = n³ + 2".

Our base case P(1) = "a₁ = 1³ + 2", which is true because a₁ is defined to be 3 and the right-hand side evaluates to 3.

For the inductive step, we assume a_n-1 = (n-1)² + 2 and try to prove that a_n = n³ + 2.

Using the rule for the sequence, a_n is equal to a_n-1 + 3n² - 3n + 1 which is equal to (n-1)³ + 2 + 3n² - 3n + 1. Expanding the first term, this is n³ - 3n² + 3n - 1 + 2 + 3n² - 3n + 1, which simplifies to n³ + 2.

We have completed the base case and the inductive step, and thus proved that P(n) is true for all positive integers n.

Question 5 (25): The Tribonacci sequence is defined by the rules T₁ = 1, T₂ = 1, T₃ = 1, and for all n with n > 3, T_n = T_n-1 + T_n-2 + T_n-3.

(a, 5) Compute T_k for all k such that 1 ≤ k ≤ 7.
T₁, T₂, and T₃ are all 1 as given. T₄ = 1 + 1 + 1 = 3, T₅ = 1 + 1 + 3 = 5, T₆ = 1 + 3 + 5 = 9, and T₇ = 3 + 5 + 9 = 17.
(b, 20) Prove by induction that for all positive integers n, T(n) is odd. You will need separate base cases for n = 1, n = 2, and n = 3.
Our statement P(n) is "T_n is odd".
For the three base cases, we just need to observe that T₁, T₂, and T₃ are each odd numbers since they are each 1.
For the inductive case, we assume that P(i) is true for all i with 1 ≤ i ≤ n - 1 and use that assumption to prove P(n).
If n > 3, then n - 1, n - 2, and n - 3 are each positive integers and we may assume that T_n-1, T_n-2, and T_n-3 are each odd numbers. If we write them as 2a + 1, 2b + 1, and 2c + 1 respectively, we can see that T_n = 2(a + b + c) + 3 and is thus an odd number itself.
Some people said that each of the three odd numbers was equal to the same number 2k + 1, whereas all we may assume about them is that each is odd.
One person had an interesting alternate approach, expanding T_n-1 to get that T_n = (T₂ + T₃ + T₄) + T₂ + T₃, which is 2(T₂ + T₃) + T₄ and thus a multiple of two plus a number assumed to be odd, and hence an odd number. This is a correct proof except for one problem. It asssumes that T_n-4 is odd, which we only know if n - 4 is a positive integer, which is only true for n ≥ 5. Since we have only assumed that n ≥ 4, to use this approach we have to deal with the case of n = 4 separately, referring to part (a) and observing that 3 is odd.

Last modified 25 October 2015